WKB Connection Formulas

For $z \gg 0$, Airy’s function gets a nice approximation:

$$\psi_{p}(x) \;=\; aAi(z) \;+\; bBi(z)$$$$\begin{cases}{Ai(z) \;\cong\; \dfrac{1}{2\sqrt{\pi}\,z^{\frac{1}{4}}}e^{-\frac{2}{3}z^{\frac{2}{3}}}}\\{\quad}\\{Bi(z) \;\cong\; \dfrac{1}{\sqrt{\pi}\,z^{\frac{1}{4}}}e^{\frac{2}{3}z^{\frac{2}{3}}}}\end{cases}$$

OK so let’s smoosh that into a single line:

$$\psi_{p}(x) \;=\; \frac{a}{2\sqrt{\pi}\,z^{\frac{1}{4}}}e^{-\frac{2}{3}z^{\frac{2}{3}}} \;+\; \frac{b}{\sqrt{\pi}\,z^{\frac{1}{4}}}e^{\frac{2}{3}z^{\frac{2}{3}}} \\ \;=\; \frac{a}{2\sqrt{\pi}\left(\alpha x\right)^{\frac{1}{4}}}e^{-\frac{2}{3}z^{\frac{2}{3}}} \;+\; \frac{b}{\sqrt{\pi}\left(\alpha x\right)^{\frac{1}{4}}}e^{\frac{2}{3}z^{\frac{2}{3}}}$$

And this guy right here —

$$\psi_{WKB} \;=\; \frac{D}{\sqrt{\hbar}\,\alpha^{\frac{3}{4}}x^{\frac{1}{4}}}e^{-\frac{2}{3}\left(\alpha x\right)^{\frac{3}{2}}}$$

— is supposed to be the same thing as that guy.

Which means $b=0$. The growing exponential has to die.

$$\frac{a}{2\sqrt{\pi}\left(\alpha x\right)^{\frac{1}{4}}}e^{-\frac{2}{3}z^{\frac{2}{3}}} \;=\; \frac{D}{\sqrt{\hbar}\,\alpha^{\frac{3}{4}}x^{\frac{1}{4}}}e^{-\frac{2}{3}\left(\alpha x\right)^{\frac{3}{2}}}\\ D \;=\; \frac{a\sqrt{\hbar}\,\alpha^{\frac{3}{4}}x^{\frac{1}{4}}}{2\sqrt{\pi}\,\alpha^{\frac{1}{4}}x^{\frac{1}{4}}} \;=\; \frac{a\sqrt{\hbar}}{2\sqrt{\pi}}\alpha^{\frac{1}{2}} \;=\; \frac{a}{2}\sqrt{\frac{\hbar\alpha}{\pi}} \\ \therefore\; a \;=\; \sqrt{\frac{4\pi}{\alpha\hbar}}\,D$$

Boom — we’ve got $a$ and $D$ locked together!!

Now let’s run the exact same play on the $x<0$ side.

$$\psi_{WKB} \;=\; \frac{1}{\sqrt{p(x)}}\left[B e^{\frac{i}{\hbar}\int_{x}^{0}p(x')dx'} \;+\; C e^{-\frac{i}{\hbar}\int_{x}^{0}p(x')dx'}\right]$$

First, pin down $p(x)$. And the integral of $p(x)$ while we’re at it!

$$p(x)=\sqrt{2m(E-V(x))}\quad \text{— plug in the linear approximation of }V(x)\\ p(x) \;=\; \sqrt{2m(E-E-V'(0)x)} \;=\; \sqrt{-2mV'(0)x} \;=\; \sqrt{-\hbar^{2}\alpha^{3}x} \;=\; \hbar\alpha^{\frac{3}{2}}\sqrt{-x} \;=\; \hbar\alpha^{\frac{3}{2}}\left(-x\right)^{\frac{1}{2}}$$

Now the integral:

$$\int_{x}^{0}p(x')dx' \;=\; \int_{x}^{0}\hbar\alpha^{\frac{3}{2}}\left(-x'\right)^{\frac{1}{2}}dx' \;=\; \hbar\alpha^{\frac{3}{2}}\int_{x}^{0}\left(-x'\right)^{\frac{1}{2}}dx' \;=\; \frac{2}{3}\hbar\left(-\alpha x\right)^{\frac{3}{2}}$$

So,

$$\psi_{WKB} \;=\; \frac{1}{\sqrt{p(x)}}\left[B e^{\frac{i}{\hbar}\int_{x}^{0}p(x')dx'} \;+\; C e^{-\frac{i}{\hbar}\int_{x}^{0}p(x')dx'}\right] \\ \;=\; \frac{1}{\sqrt{\hbar}\,\alpha^{\frac{3}{4}}(-x)^{\frac{1}{4}}}\left[B e^{i\frac{2}{3}\left(-\alpha x\right)^{\frac{3}{2}}} \;+\; C e^{-i\frac{2}{3}\left(-\alpha x\right)^{\frac{3}{2}}}\right]$$

And what is THIS the same as?!

$$\psi_{p}(x) \;=\; aAi(z) \;+\; bBi(z)$$

Yep. Same thing.

But — same move as before — we don’t use the “near $z=0$” form of the Airy functions. That one’s only good right around $z\sim 0$. For $z \ll 0$ we need the oscillatory asymptotic form:

$$\psi_{p}(x) \;=\; aAi(z) \;+\; bBi(z)$$$$\begin{cases}{Ai(z) \;=\; \dfrac{1}{\sqrt{\pi}(-z)^{\frac{1}{4}}}\sin\left(\dfrac{2}{3}\left(-z\right)^{\frac{3}{2}} + \dfrac{\pi}{4}\right)}\\{\quad}\\{Bi(z) \;=\; \dfrac{1}{\sqrt{\pi}(-z)^{\frac{1}{4}}}\cos\left(\dfrac{2}{3}\left(-z\right)^{\frac{3}{2}} + \dfrac{\pi}{4}\right)}\end{cases}$$

Now, time to say these two are the same.

Remember we already got $b=0$? Yeah, $b=0$ here too, obviously — it’s still the same patching function. We’re using one $\psi_p$ to bridge both sides.

So the patching function on $x<0$ is

$$\psi_{p} \;\cong\; \frac{a}{\sqrt{\pi}(-z)^{\frac{1}{4}}}\sin\left(\frac{2}{3}(-z)^{\frac{3}{2}} + \frac{\pi}{4}\right)$$

Problem: this is a $\sin$. The WKB form has exponentials. Can’t match directly.

So I’ll call in Uncle Euler and blow the $\sin$ apart into exponentials:

$$\psi_{p} \;\cong\; \frac{a}{\sqrt{\pi}(-z)^{\frac{1}{4}}}\sin\left(\frac{2}{3}(-z)^{\frac{3}{2}} + \frac{\pi}{4}\right) \\ =\; \frac{a}{\sqrt{\pi}(-z)^{\frac{1}{4}}}\left(\frac{e^{i\left(\frac{2}{3}(-z)^{\frac{3}{2}} + \frac{\pi}{4}\right)} - e^{-i\left(\frac{2}{3}(-z)^{\frac{3}{2}} + \frac{\pi}{4}\right)}}{2i}\right) \\ =\; \frac{a}{\sqrt{\pi}(-z)^{\frac{1}{4}}}\frac{1}{2i}\left(e^{i\frac{\pi}{4}}e^{i\frac{2}{3}(-z)^{\frac{3}{2}}} - e^{-i\frac{\pi}{4}}e^{-i\frac{2}{3}(-z)^{\frac{3}{2}}}\right)$$

Set this equal to

$$\psi_{WKB} \;=\; \frac{1}{\sqrt{\hbar}\,\alpha^{\frac{3}{4}}(-x)^{\frac{1}{4}}}\left[B e^{i\frac{2}{3}\left(-\alpha x\right)^{\frac{3}{2}}} \;+\; C e^{-i\frac{2}{3}\left(-\alpha x\right)^{\frac{3}{2}}}\right]$$

match exponentials with exponentials, and out pops the relationship between $B, C$ and $a$:

$$B \;=\; \frac{\sqrt{\hbar\alpha}}{2i\sqrt{\pi}}e^{i\frac{\pi}{4}}a\\ C \;=\; -\frac{\sqrt{\hbar\alpha}}{2i\sqrt{\pi}}e^{-i\frac{\pi}{4}}a$$

WAIT WAIT WAIT. We already had

$$a \;=\; \sqrt{\frac{4\pi}{\alpha\hbar}}D$$

from the other side!!!

So $D$ is the bridge — sub it in and $B, C, D$ are all holding hands:

$$B \;=\; -ie^{i\frac{\pi}{4}}D\\ C \;=\; ie^{-i\frac{\pi}{4}}D$$

Now plug those back into the WKB for $x<0$:

$$\psi_{WKB} \;=\; \frac{1}{\sqrt{p(x)}}\left[B e^{\frac{i}{\hbar}\int_{x}^{0}p(x')dx'} \;+\; C e^{-\frac{i}{\hbar}\int_{x}^{0}p(x')dx'}\right] \\ \;=\; \frac{D}{\sqrt{p(x)}}\left(-ie^{i\frac{\pi}{4}}e^{i\frac{1}{\hbar}\int_{x}^{0}p(x')dx'} \;+\; ie^{-i\frac{\pi}{4}}e^{-i\frac{1}{\hbar}\int_{x}^{0}p(x')dx'}\right)$$

Clean it up:

$$=\; \frac{-iD}{\sqrt{p(x)}}\left(e^{i\left(\frac{1}{\hbar}\int_{x}^{0}p(x')dx' + \frac{\pi}{4}\right)} - e^{-i\left(\frac{1}{\hbar}\int_{x}^{0}p(x')dx' + \frac{\pi}{4}\right)}\right) \\ =\; \frac{-iD}{\sqrt{p(x)}}(2i)\frac{\left(e^{i\left(\frac{1}{\hbar}\int_{x}^{0}p(x')dx' + \frac{\pi}{4}\right)} - e^{-i\left(\frac{1}{\hbar}\int_{x}^{0}p(x')dx' + \frac{\pi}{4}\right)}\right)}{2i}\\ =\; \frac{2D}{\sqrt{p(x)}}\sin\left(\frac{1}{\hbar}\int_{x}^{0}p(x')dx' + \frac{\pi}{4}\right) \\ \\ \therefore\; \psi_{WKB} \;=\; \frac{2D}{\sqrt{p(x)}}\sin\left(\frac{1}{\hbar}\int_{x}^{0}p(x')dx' + \frac{\pi}{4}\right)$$

And there it is — we’ve patched WKB so it doesn’t choke at $x=0$!!

We took the potential near the turning point where $V\approx E$, used that to fix WKB’s blowup, and the connection formula fused the bound-state side and the tunneling side — two previously unrelated pieces — into one:

$$\psi_{WKB} \;=\; \begin{cases}{\dfrac{2D}{\sqrt{p(x)}}\sin\left(\dfrac{1}{\hbar}\int_{x}^{0}p(x')dx' + \dfrac{\pi}{4}\right) \quad (x<0)}\\{\quad}\\{\quad}\\{\dfrac{D}{\sqrt{\left|p(x)\right|}}e^{-\frac{1}{\hbar}\int_{0}^{x}\left|p(x')\right|dx'} \quad (x>0)}\end{cases}$$

And the one remaining mystery — how do we actually pin down $D$?

Normalization. That’s it. Normalize the thing and $D$ falls out.

And we’re done ~~

---

Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.